The one saith,
This is my son that liveth, and thy son is the dead; and the other saith,
Nay, thy son is the dead, and my son is the living. And the king said, Bring
me a sword.

1 Kings 3:23

Suppose
a rectangular cake has special decorative frosting on one rectangular region,
and we wish to cut the cake along a single straight line in such a way that each
side contains exactly half of the cake, and also half of the special
frosting. This is accomplished by cutting along the line through the
geometric centers of the two rectangular regions, as shown below.

This
works because every rectangle is an “even” shape, in the sense that the
radial distance of the perimeter from the centroid satisfies r(θ) =
r(-θ). As a result, every line through the centroid cuts the rectangle
into two equal areas. The same applies to regular hexagons, octagons,
ellipses, etc., but not to triangles, pentagons, or other “odd” shapes. (By
the way, since the frosting is only on the top surface, this problem
precludes the wise-guy solution of slicing the cake along the plane half-way
between the top and bottom of the cake.)

The
radio show “Car Talk” posed a somewhat similar puzzle in 2012 involving a rectangular
tray of brownies with a rectangular piece cut out of the interior, as
illustrated below.

The
question posed was “How can you, with one cut of the knife, cut the brownies
in half?” In this puzzle there is no stated requirement for each “part”
to contain an equal amount of the cutout region, so there are actually
infinitely many different straight lines that bisect the area of the brownies
(as explained below). Also, the stated question didn’t actually specify that
the cut must be along a straight line, and in addition it’s not entirely
clear what is meant by “one cut of the knife”. If we cut along a line that
begins at the edge of the tray and passes through the empty region and then
continues (along the same line) through another region of the brownie until
reaching the edge of the tray, is this considered one cut or two? Suppose the
interior rectangle was nearly as large as the entire tray, so the remaining brownie
consists of just a narrow rectangular path along the edge of the tray, and we
make cuts on opposite sides, as shown below.

Would
this be considered one cut, or two? Granted, both cuts are along the same
straight line, but the question did not ask for two cuts along a single
straight line, it asked for a “single cut”, and not necessarily straight. One
could argue that a “single cut” must consist of a single unbroken cut, so the
only solution (in this case) would be to cut around the perimeter. The same approach
could be applied in the general case, as illustrated below.

To
draw the cut line we arbitrarily selected a point P inside the cutout region
and plotted all the points at a distance rc from that point, where
rc is the root mean square of the distances (along the same ray)
from P to the inner and outer boundaries, denoted by r1 and r2
in the figure above. This ensures that the incremental areas swept out by the
ray rotating about the point P inside and outside the cut line are equal.
This follows from the fact that the incremental triangular area swept out by
a segment of length r rotating by an incremental angle dθ about the
point P is (r/2)rdθ, and so equating the areas inside and outside rc
results in the condition

and
therefore

Thus
for any choice of the point P we get a single continuous cut that splits the
brownie into two equal areas. This is arguably the only acceptable answer to
the Car Talk puzzle as stated (at least the only generally applicable one,
for any size of the interior cutout), since the question required a “single
cut”, and of course it did not specify that the single cut had to be along a
straight line.

However,
one suspects that the question Car Talk intended to ask was “How can you,
by cutting along a single straight line, cut the brownies in half?” This
differs from the question that was actually asked in two respects. First, it
eliminates the problematic requirement for a “single cut”, and second, it
adds the requirement for the cutting to all lie along a single straight line.
It still differs, though, from our original question about cutting the cake,
because (as noted above) it doesn’t specify that each part gets an equal
amount of the cutout region. Without that stipulation, there are obviously an
infinite number of straight lines the bisect the brownies into two equal
areas. To see this, consider a line, with fixed orientation, sweeping across
the tray, and note that all of the area will initially be on one side of the
line, and then, after the line has swept all the way across the tray, all the
area will be on the other side of the line. The area on one side is
continuously increasing as we slide the line, so there must be a position
where exactly half the area has been passed. This applies to any orientation
of the line, so we can cut the brownies in half along a line at any angle we
choose. In fact, this same reasoning applies to more general shapes, including
triangles, etc.

We
might ask whether, for a triangular cake with a triangular region of special
decorative frosting, there exists a straight line that divides both the outer
and the inner triangle into equal areas (analogous to the line through the
centroids of rectangles). Note that the line need not pass through the
centroids, but it must bisect each of the triangles into equal areas, which
is a stronger condition than simply bisecting the outer region into equal
areas. To answer this question, let’s first consider how to find the lines
that divide a given triangle into two equal areas. Consider the triangular
region shown below.

The
symbol λab is the fraction of the distance from a to b where
the point s is located. As s ranges from a to b, the value of λab
ranges from 0 to 1. The value of λac is defined similarly
based on the position of point r. We also note the area relation Aabc
= Aasr + Asbcr. Now, in terms of these parameters, and
noting the area of a triangle is half the product of any two edge lengths
times the sine of the angle between them, we have

It
follows that the area is cut into equal parts if

Now
consider a triangular cake with a triangular region with special frosting. By
moving a point along the perimeter of the outer triangle, and using the above
relation to determine the corresponding point on the next clockwise edge, we
can create the continuous sequence of lines, each of which bisects the area
of the outer triangle. These lines rotate continuously through a complete
revolution, and sweep through the entire interior of the outer triangle, as
shown in the figure below.

Hence
they sweep through any interior figure, including the internal triangle, so
we can find one of these lines that bisects the area of the internal
triangle. An example is shown in the figure below.

For
any value of λca between 0.5 and 1 we can compute the value
of λcb such that λcaλcb=1/2, which
gives a line bisecting the area of the outer triangle, and then we can
compute the points of intersection of this line with the sides of the inner
triangle, from which we get the values of λed and λef.
Our objective is to determine a value of λca such that λedλef
= 1/2. The coordinates of the vertices of the outer triangle in the figure
above are (-7,-2), (8,6), and (2,-5), and the coordinates of the vertices of
the inner triangle are (1,-1), (3,2), and (4,0). From this we can determine
that the values of λca that satisfy the stated conditions
must be a root of the quartic polynomial

The
only real root in the range from 0.5 to 1 is λca =
0.9252391... , and this is depicted in the figure above, showing that it
does indeed bisect both the overall cake and the region of special frosting.

Incidentally,
although we’ve restricted the values of the λ parameters to the range 0
to 1, so that the interpolated points are between the vertices, the same
equations apply formally to arbitrary values, based on the extrapolated edges
of the triangles. For example, in the triangle below we have chosen a value
of λab greater than 1.

Just
as before we have the area relation Aabc = Aasr + Asbcr,
with the understanding that these are signed areas, with the regions enclosed
by clockwise paths defined as positive. Thus the region stb is positive for
the triangle asr but negative for the quadrilateral sbcr. The difference
between the signed area of the quadrilateral sbcr and the triangle asr is

and
hence these two signed areas are equal if λcaλcb=1/2
for any values of the parameters, and we can say that the line rs bisects the
area of the triangle abc. On this basis, the other real root of the quartic
polynomial, λca = 0.333889..., gives another solution, albeit
purely formal, since there is no negative cake. There are also two complex
solutions.